Question: Solve for $x$ and $y$ using elimination. $\begin{align*}8x-8y &= 6 \\ 8x-2y &= -3\end{align*}$
Solution: We can eliminate $x$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-1$ and the bottom equation by $1$ $\begin{align*}-8x+8y &= -6\\ 8x-2y &= -3\end{align*}$ Add the top and bottom equations. $6y = -9$ Divide both sides by $6$ and reduce as necessary. $y = -\dfrac{3}{2}$ Substitute $-\dfrac{3}{2}$ for $y$ in the top equation. $8x-8( -\dfrac{3}{2}) = 6$ $8x+12 = 6$ $8x = -6$ $x = -\dfrac{3}{4}$ The solution is $\enspace x = -\dfrac{3}{4}, \enspace y = -\dfrac{3}{2}$.